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course Phy 242
3/15 6
`q001. A ray is directed parallel to the central axis of a circular mirror, 2 cm from the central axis. The mirror has radius 10 cm. A radial line (i.e., a line starting at the center of the mirror) through the point at which the ray strikes the mirror is normal (perpendicular) to the mirror's surface at that point. Show why the reflected ray strikes the axis at a point which is close to 5 cm from the center (and also 5 cm from the mirror). (This point is called the 'focal point' of the mirror).The ray when hitting the curved surface doesn’t know the surface is curved, so that when it hits it is reflected at whatever angle the tang plane is for that specific point causing it to redirect its motion toward the focal point.
?????This will be approx. the same for all rays that are sent parallel to the axial line right????????
@& This isn't the case if the ray is far from the axis.
You need to construct triangles, as in the Class Notes, and argue in terms of similar triangles.
What triangles appear to be similar?
None of them really are perfectly similar. However if the distance from the axis is small, there are triangles that approach similarily and allow us to establish the result.*@
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The tip of a flame 'sends out' rays in all directions (see the above figure). Assume that the flame is 2 cm high and consider two of these: the ray parallel to the axis, which reflects through the focal point, and the ray which strikes the mirror along its axis and hence reflects as if the mirror was flat instead of curved. These two rays meet at a point, as shown below. If the flame is 20 cm from the mirror, how far will that meeting point be from the mirror, and from the central axis? If the flame is moved further from the mirror, will the meeting point move further from or closer to the mirror, and will it move further from or closer to the central axis?
They will meet at approx. 20cm from mirror and 2cm from axial ray because the object dist. is greater than the focal dist. It will move closer to both mirror and central axis, if object is moved closer meeting pt will move further and further away until object dist is equal to or less than focal dist at which rays never meet on opposite side of lens.
@& The ray through the focus is 'steeper' than the ray through the center, and will meet that ray at a point closer to the lens than 20 cm.
If the flame is very far from the axis, then the reflected central ray will be nearly parallel to the axis, so that the parallel and central rays will meet very close to the focus.
If the flame is near the focus the parallel ray will be reflected almost parallel to the central ray, and the rays will intersect far away.*@
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`q002. A laser ray directed from the left parallel to the central axis of a lens, but 2 cm from the axis, is refracted by the lens and passes through the axis at a point 10 cm to the right of the lens. A second ray originates from a point 20 cm to the left of the lens, strikes the lens on the central axis and passes through with practically no deflection from its original path.
How far to the right of the lens will this ray intersect the first, and how far will it be from the axis when that occurs?
They will meet at approx. 20cm from mirror and 2cm from axial ray because the object dist. is greater than the focal dist.
???I’m not sure I understand what’s being asked because I answered this same question previously so I’m not sure which one I answered incorrectly??????
@& This case is for a lens, not a mirror.
The reasoning is the same, if you replace the word 'reflected' by 'refracted'.*@
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`q003. A laser ray enters a cylindrical lens of radius 10 cm, along a path parallel to but 2 cm from the central axis. It is deflected at that point toward a radial line, in such a way that its angle with the radial line is decreased by 50%. Will the ray get to the opposite surface of the mirror before it reaches the central axis? Where will it go then?
Sin(`theta_i) = 2cm/10cm = .2, taking 50%of that we get .1
`theta_r = .1, also has the same adjacent side = 2cm
so
Sin(`theta_r) = 2cm/x = .1
x = 20cm, so that for `theta_r 50% less than `theta_i it will hit right at the opposite side of circle, if center of circle represents ordered pair (0, 0) then we expect `theta_r to cross x axis at (10, 0). This is for small values of `dx because basing everything on the triangle formed by `dx and R we are not accounting for small distance between base of triangle side `dx and outer most edge of circle. The larger `dx gets the larger the dist between base of triangle and outer most edge of circle.
@& Very good.
Another way to see that this isn't exact: The sine function is not quite linear, though for these angles it is very nearly so. If you split hairs you will find that the sine of half the angle is a little more than half the sine of the angle, but the difference is slight. (consider the Taylor approximation sin(x) = x - x^3 / 3 ! + ... ; x^3 is much bigger for the larger angle)*@
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`q004. I send two waves with wavelengths 12 cm are sent down two chains, held in my hands like the reins of a horse. At some distance from my hands the two chains meet a third chain at a common point, forming a narrow Y. I create the two waves by moving my hands in opposite directions, alternately moving them further apart and closer together. The waves travel at the same speed toward their meeting point and travel the same distance before they get there.
Will this create a wave in the third chain?
Yes because nodes and antinodes are running concurrent together.
@& One chain will transmit a pulse to the right, the other to the left.
The two pulses, running the same distance, will therefore have opposite effects and cancel.
Note that while I can interpret the meaning, 'nodes' and 'antinodes' are terms that apply to standing waves, not to traveling waves.*@
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If I now move my left hand 6 cm closer to the point where the chains meet, will a wave be created in the third chain?
Path difference of 6cm is (1/2) wavelength difference so there waves just won’t be perfectly in phase causing there to appear to be no waves.
????It is only when path difference is (1/2) wavelength waves will arrive outta phase and cancel or any mult. of (1/2) like 1.5, 2, 2.5, 3, etc. Does that mean you observe absolutely no waves as they are traveling down string. I tried to recreate this with my sack of rubber bands but can’t get it right???
@& Good reasoning, but the key is that these pulses start out 180 deg out of phase, because my hands are moving apart. The pulse from the closer of the two chains will be unopposed. The pulse in that chain will be on the left when the pulse from the further chain arrives. This pulses and the original pulse from the further chain will therefore arrive at the same time and reinforce. This will be followed by alternating rightward and leftward reinforcing pulses, creating a wave in the chain.*@
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If without changing my grip I pull my left hand back to its original position, stretching the chain so the distances to the third chain are now equal, and repeat the same motion, how might this affect what happens in the third chain?
Waves will be moving in and out of phase with each other causing there to be waves at some points on the third string but not consistently.
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@& The tension and therefore the propagation velocity will change when I pull the chain back. You would need further information (e.g., the relationship between length and tension, and the mass density) to determine exactly how this affect the phase relationship.*@
&#This looks good. See my notes. Let me know if you have any questions. &#